试图让Froala将图像上传到PHP环境中的服务器。完全遵循Froala的例子here。
但是，当我选择一个图像文件并提交时，我收到的反馈是一条错误消息：“出错了。请再试一次”。
我已经使用了示例中的确切文件/目录名称，但显然我遗漏了一些东西。该目录是图像存储库的“上传”。
我已经阅读了大多数其他Stackoverflow评论和解决方案，尝试过其中几个，阅读Froala的信息，但仍然没有成功。
我正在使用的代码与他们的示例完全相同。
编辑器文件：“index.php”
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<!-- Include external CSS. -->
<link href="https://cdnjs.cloudflare.com/ajax/libs/font-awesome/4.4.0/css/font-awesome.min.css" rel="stylesheet" type="text/css" />
<link rel="stylesheet" href="https://cdnjs.cloudflare.com/ajax/libs/codemirror/5.25.0/codemirror.min.css">
<!-- Include Editor style. -->
<link href="https://cdnjs.cloudflare.com/ajax/libs/froala-editor/2.5.1/css/froala_editor.pkgd.min.css" rel="stylesheet" type="text/css" />
<link href="https://cdnjs.cloudflare.com/ajax/libs/froala-editor/2.5.1/css/froala_style.min.css" rel="stylesheet" type="text/css" />
</head>
<body>
<!-- Include external JS libs. -->
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/jquery/1.11.0/jquery.min.js"></script>
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/codemirror/5.25.0/codemirror.min.js"></script>
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/codemirror/5.25.0/mode/xml/xml.min.js"></script>
<!-- Include Editor JS files. -->
<script type="text/javascript" src="https://cdnjs.cloudflare.com/ajax/libs/froala-editor/2.5.1//js/froala_editor.pkgd.min.js"></script>
<div class="sample">
<h2>File upload example.</h2>
<form>
<textarea id="edit" name="content"></textarea>
</form>
</div>
<!-- Initialize the editor. -->
<script>
$(function() {
$('#edit').froalaEditor({
// Set the file upload URL.
imageUploadURL: '/upload_image.php',
imageUploadParams: {
id: 'my_editor'
}
})
});
</script>
</body>
</html>
PHP文件：“upload_image.php”
<?php
try {
// File Route.
$fileRoute = "/uploads/";
$fieldname = "file";
// Get filename.
$filename = explode(".", $_FILES[$fieldname]["name"]);
// Validate uploaded files.
// Do not use $_FILES["file"]["type"] as it can be easily forged.
$finfo = finfo_open(FILEINFO_MIME_TYPE);
// Get temp file name.
$tmpName = $_FILES[$fieldname]["tmp_name"];
// Get mime type.
$mimeType = finfo_file($finfo, $tmpName);
// Get extension. You must include fileinfo PHP extension.
$extension = end($filename);
// Allowed extensions.
$allowedExts = array("gif", "jpeg", "jpg", "png", "svg", "blob");
// Allowed mime types.
$allowedMimeTypes = array("image/gif", "image/jpeg", "image/pjpeg", "image/x-png", "image/png", "image/svg+xml");
// Validate image.
if (!in_array(strtolower($mimeType), $allowedMimeTypes) || !in_array(strtolower($extension), $allowedExts)) {
throw new \Exception("File does not meet the validation.");
}
// Generate new random name.
$name = sha1(microtime()) . "." . $extension;
$fullNamePath = dirname(__FILE__) . $fileRoute . $name;
// Check server protocol and load resources accordingly.
if (isset($_SERVER["HTTPS"]) && $_SERVER["HTTPS"] != "off") {
$protocol = "https://";
} else {
$protocol = "http://";
}
// Save file in the uploads folder.
move_uploaded_file($tmpName, $fullNamePath);
// Generate response.
$response = new \StdClass;
$response->link = $protocol.$_SERVER["HTTP_HOST"].dirname($_SERVER["PHP_SELF"]).$fileRoute . $name;
// Send response.
echo stripslashes(json_encode($response));
} catch (Exception $e) {
// Send error response.
echo $e->getMessage();
http_response_code(404);
}
?>
来自他们的网站。
在进行任何上传之前，必须将uploads目录设置为有效位置。该路径可以是任何可访问和可写的文件夹。
请检查您的文件夹权限，这是一个常见的疏忽。
问候
edwoli
通过执行以下操作使其工作：
imageUploadURL: '/upload_image.php',更改为imageUploadURL: 'upload_image.php',很难相信Froala网站示例中有这个错误，并且没有得到纠正......？编辑：Froala评论说，如果文件在根目录中，“/”将起作用。我不是。